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Sloan [31]
2 years ago
10

A student borrows $10,000 from a bank but ends up paying $13,000 on the

Chemistry
1 answer:
NeTakaya2 years ago
4 0

The student paid an interest of $3,000 over the next five years.

HOW TO CALCULATE INTEREST:

The interest on a borrowed amount of money can be calculated by subtracting the principal from the amount paid over time. That is;

Interest = principal - amount

According to this question, a student borrows $10,000 from a bank but ends up paying $13,000. This means that the principal is $10,000 while the amount is $13,000. The interest is calculated thus;

Interest = $13,000 - $10,000

Interest = $3,000

Therefore, the student paid an interest of $3,000 over the next five years.

Learn more about interest at: brainly.com/question/4626564

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(a) if a sample containing 2.00 ml of nitroglycerin is detonated, how many total moles of gas are produced? (b) if each mole of
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Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:P4(g) + 10 Cl2(g) → 4PCl5(s) ΔH°
miskamm [114]

<u>Answer:</u> The \Delta H^o_{rxn} for the reaction is -1835 kJ.

<u>Explanation:</u>

Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction follows:

P_4(g)+10Cl_2(g)\rightarrow 4PCl_5(s)      \Delta H^o_{rxn}=?

The intermediate balanced chemical reaction are:

(1) PCl_5(s)\rightarrow PCl_3(g)+Cl_2(g)    \Delta H_1=157kJ   ( × 4)

(2) P_4(g)+6Cl_2(g)\rightarrow 4PCl_3(g)     \Delta H_2=-1207kJ

The expression for enthalpy of the reaction follows:

\Delta H^o_{rxn}=[4\times (-\Delta H_1)]+[1\times \Delta H_2]

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(4\times (-157))+(1\times (-1207))=-1835kJ

Hence, the \Delta H^o_{rxn} for the reaction is -1835 kJ.

6 0
2 years ago
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